A carrier wave with amplitude of 250 V is amplitude modulated by a sinusoidal base band signal of amplitude 150 V. The ratio of minimum amplitude to maximum amplitude for the amplitude modulated wave is 50 : $$x$$, then value of $$x$$, is _________.

We start by recalling the standard expression for a sinusoidally amplitude-modulated (AM) signal. If the unmodulated carrier voltage is $$v_c(t)=A_c\cos\omega_ct$$ and the modulating or base-band signal is $$v_m(t)=A_m\cos\omega_mt$$, then the AM wave is written as

$$v_{\text{AM}}(t)=A_c\left(1+m\cos\omega_mt\right)\cos\omega_ct.$$

Here $$A_c$$ is the carrier amplitude, $$A_m$$ is the message (modulating) amplitude and $$m$$ is the modulation index, defined by the formula

$$m=\frac{A_m}{A_c}.$$

In the given problem the carrier amplitude is $$A_c=250\text{ V}$$ and the modulating amplitude is $$A_m=150\text{ V}.$$ Using the definition, we calculate

$$m=\frac{A_m}{A_c}=\frac{150}{250}=0.6.$$

The envelope (overall amplitude) of the AM wave varies between a maximum value $$A_{\text{max}}$$ and a minimum value $$A_{\text{min}}$$. These are obtained by substituting the extreme values of $$\cos\omega_mt$$, namely $$+1$$ and $$-1$$, into the factor $$(1+m\cos\omega_mt)$$:

$$A_{\text{max}}=A_c(1+m), \qquad A_{\text{min}}=A_c(1-m).$$

Substituting $$A_c=250\text{ V}$$ and $$m=0.6$$ gives

$$A_{\text{max}}=250(1+0.6)=250\times1.6=400\text{ V},$$

$$A_{\text{min}}=250(1-0.6)=250\times0.4=100\text{ V}.$$

The question states that the ratio of minimum amplitude to maximum amplitude is $$50:x$$. Numerically, we have

$$\frac{A_{\text{min}}}{A_{\text{max}}}=\frac{100}{400}=\frac{1}{4}.$$

This must equal the given ratio $$\dfrac{50}{x}$$. Hence we write

$$\frac{50}{x}=\frac{1}{4}\quad\Longrightarrow\quad 50\cdot4=x\quad\Longrightarrow\quad x=200.$$

So, the answer is $$200$$.